# Questions about the dimension-and other properties-of a non-separable topological space

Let the topological space X be the so-called "long line" — which is an uncountable linearly ordered set containing all the countable ordinal numbers as well as a copy of the open unit interval (0,1) between each countable ordinal number and its successor. The topology of X is its order topology. Let the topological space Y be the 2-dimensional sphere-more precisely, let Y be the boundary of the closed unit ball in 3-dimensional Euclidean space. Let Z be the topological product of X and Y.

My questions are:

1. Is Z 3-dimensional, and if not, what is its dimension?
2. Does Z contain an uncountable collection of pairwise disjoint open subsets?

I mean by "dimension", the so-called "small inductive dimension". This is a well known function, which for any non-empty regular Hausdorff space (as argument) has a non-negative integer value (or is infinite). The function has the same value for any pair of homeomorphic spaces in its range of definition. This includes Z-which is regular and Hausdorff, since both X and Y are.

The motivation for my questions is this: The physical space of our universe is most simply and clearly pictured as a 3-dimensional Euclidean space E(3) extending to infinity in all directions-in other words, Newtonian space, before the complications of relativity theory arose. Now E(3) contains an uncountable infinity of points but can only contain a countable infinity of "disjoint discrete objects" such as stars, galaxies or even atoms.

What simply described topological space could serve as the space of a physical universe containing uncountably many "disjoint discrete objects"? Such a space should probably contain an uncountable set of pairwise disjoint open subsets. A non-separable Hilbert space would meet this condition but its dimension is infinite. Could there not exist a low dimensional space, closer to E(3), which also meets this condition? That is why I am asking these questions about the arc-wise connected space Z. I also wonder whether Z shares the following desirable property of Euclidean spaces-being homogeneous and isotropic. If p and q are distinct points of Z, does there always exist a homeomorphism of Z onto itself that takes p into q?

• Could you divide your question into paragraphs? It's a bit hard to read as a single block. Oct 7, 2014 at 21:10
• For your third question, $Z$ is a manifold and any manifold is homogeneous. Oct 8, 2014 at 7:24
• I would like to break up that whole pile of verbosity into Oct 9, 2014 at 20:13
• I would like to break up that whole pile of verbosity into separate sections or paragraphs but do not know how to do this. I tried several times but each attempt caused me to lose everything that I had written previously. Oct 9, 2014 at 20:25

1. Dimension is a local property (there is a possibility that I am mistaken, please confirm) and the long line is locally homeomorphic to the real line, so $X$ has dimension 1. By the same argument $Z$ has dimension 3.
2. Yes, and the open sets can even have pairwise disjoint closures. (More precisely, the closure of each open set will have a neighborhood that does not meet the closure of any other set in the collection.) It suffices to find an uncountable collection of open sets on $X$ with pairwise disjoint closures, since their products with $Y$ will give a similar collection for $Z$. Recall that $X=\omega_1\times[0,1)$. Let $A=\{\alpha+1\in\omega_1;\alpha\text{ is a limit ordinal}\}$. Now the open sets $\{\alpha\}\times(0,1)\subset X$, $\alpha\in A$, have disjoint closures and $A$ is uncountable.
• Notice that there are other non-metrizable $3$-manifolds which have the property of having pairwise disjoint open subsets, and even better: a discrete such collection. An example is $P\times\mathbb{R}$, where $P$ is any variant of the Prüfer surface (look for it in Wikipedia, for instance). And, again, any manifold is homogeneous. Oct 13, 2014 at 18:37